Prove that the following sequences monotonically decrease and increase correspondingly. Since they are bounded, find the limit.

Question

Let $\xi_n$ be a Poisson random variable with $\lambda = n \in \mathbb{N}$. That is $P(\xi_n = k) = \frac{n^ke^{-n}}{k!}$, for $k \in \mathbb{N}_0$. Let $f_+(n) = P(\xi_n \geq n),\ f_-(n) = P(\xi_n > n)$. Show that $f_+(n)$ monotonically decreases and $f_-(n)$ monotonically increases.

I have no idea how to prove this, I've tried mathematical induction and it didn't make any sense.

All what I've succeeded in is that $f_+(n) = f_-(n) + \frac{n^n e^{-n}}{n!}$. Last term here approaches $0$ when $n$ approaches infinity, what means that the limit of these two will be the same.

Answer 1

Edit: There is unfortunately a major problem as pointed out by @Davide Giraudo in the comments. For the moment, I'll leave the answer up in the hope that it can be fixed.

I have not found an easy way to get there, but here is my approach. First, take $X_i$ to be iid Poisson with parameter 1, so that we may take $\xi_n = \sum_{i=1}^n X_i$. Next, we may write \begin{align*} f_+(n+1) &= \mathbb{P}\left(X_{n+1} \geq n+1 - \sum_{i=1}^n X_i\right)\\ &= \sum_{k=0}^n \mathbb{P}(X_{n+1} \geq n+1 - k)\mathbb{P}\left(\sum_{i=1}^n X_i = k\right) + \mathbb{P}\left(\sum_{i=1}^n X_i \geq n+1\right)\\ &= \dfrac{1}{e}\sum_{k=0}^n \dfrac{1}{(n+1-k)!}\cdot \dfrac{e^{-n}n^k}{k!} + f_+(n) - \mathbb{P}\left(\sum_{i=1}^n X_i = n\right)\\ &= f_+(n) + \dfrac{1}{e}\cdot \dfrac{e^{-n}}{(n+1)!}\sum_{k=0}^n \binom{n+1}{k}n^j - \dfrac{e^{-n}n^n}{n!}\\ &= f_+(n) + \dfrac{e^{-n}}{e(n+1)!}\left( (n+1)^{n+1} - n^{n+1} - en^n(n+1)\right). \end{align*} In particular $f_+$ is decreasing if the last term in parentheses is negative. Set $$ g(x) := \dfrac{(x+1)^{x+1}}{x^x(x + e(x+1))} $$ for $x > 0$ and compute (e.g. with WolframAlpha if your are as lazy as I am) $$ g'(x) = -\dfrac{\left(\frac{1}{x} + 1\right)^x (x + 1) \cdot \Big((e x + x + e) \ln\left(\frac{x}{x + 1}\right) + e + 1\Big)}{(e x + x + e)^2}. $$ Next, we dtermine the sign of $g'$ by first nothing that, except for the last parenthesis in the nominator, everything is strictly positive, so that it suffices to note that $$ h: x\mapsto (e x + x + e) \ln\left(\frac{x}{x + 1}\right) + e + 1 $$ is a non-decreasing function (as a sum and product of non-decreasing functions). Now, as $x\to+\infty$, we have $$ h(x) \approx e\ln(0) + (1+e)\ln\left(\left(1 - \frac{1}{x+1}\right)^{x}\right) + 1 + e \approx (1+e)\ln\left(e^{-1}\right) + (1+e) = 0, $$ so that we can conclude that $h(x) \leq 0$ for all $x\geq 0$. This in turn means that $g'(x)\geq 0$ for $x\geq 0$ and therefore $g$ is non-decreasing. Again, we take a look at the limit for $x\to+\infty$ of $$ g(x) = \left(1 + \dfrac{1}{x}\right)^x\cdot \dfrac{1 + x}{(1 + e)x + e} \approx e\cdot \dfrac{1}{1+e}. $$ With this, we may conclude that $$ g(x) \leq \dfrac{e}{1+e} < 1, $$ which concludes the (extremely involved) argument.

I guess that a similar argument shows that $f_-$ is increasing, but I am not motivated enough to try it out. Also, I hope that there is an easier solution, because this is not very elegant ^^.

Answer 2

If $Y_{\lambda}$ has a Poisson distribution with parameter $\lambda>0$, then for each $n$,

$$ \tag{*} \mathbb P(Y_\lambda\leqslant n)=\int_{\lambda}^\infty \frac{t^n}{n!}e^{-t}dt. $$ This can be shown for instance by induction on $n$, where the induction step can be done via an integration by part. Letting $a_n:=1-f_+(n)=\mathbb P(\xi_n\leqslant n-1)$ and $b_n:=1-f_-(n)=\mathbb P(\xi_n\leqslant n)$, we will show that $(a_n)$ is increasing and $(b_n)$ is decreasing. Using (*), we find that \begin{align} a_{n+1}-a_n&=\mathbb P(\xi_{n+1}\leqslant n)-\mathbb P(\xi_n\leqslant n-1)\\ &=\int_{n+1}^\infty\frac{t^{n}}{n!}e^{-t}dt-\int_{n}^\infty\frac{t^{n-1}}{(n-1)!}e^{-t}dt\\ &=\int_{n+1}^\infty\frac{t^{n}}{n!}e^{-t}dt-\left(\left[\frac{t^{n}}{n!}e^{-t}\right]_{t=n}^\infty-\int_n^\infty\frac{t^{n}}{n!}(-e^{-t})dt\right)\\ &=\int_{n+1}^\infty\frac{t^{n}}{n!}e^{-t}dt+\frac{n^{n}}{n!}e^{-n}-\int_n^\infty\frac{t^{n}}{n!} e^{-t} dt\\ &=\frac{n^{n}}{n!}e^{-n}-\int_n^{n+1}\frac{t^{n}}{n!} e^{-t}dt. \end{align} Doing the substitution $s=t-n$ reduces us to show that $$ 1>\int_0^1\left(1+\frac{s}n\right)^ne^{-s}ds, $$ which follows from $\ln(1+u)< u$ for $u>0$.

By similar considerations, we find that the sign of $b_{n+1}-b_n$ is the same as that of $$ 1-\int_0^1\left(1+\frac{s}n\right)^{n+1}e^{-s}ds $$ and having a look at the function $g\colon s\mapsto (n+1)\ln(1+s/n)-s$ gives what we want.

Notice that $\xi_n$ has the same distribution as $\sum_{i=1}^n X_i$, where $(X_i)$ is an i.i.d. sequence and $X_i$ has a Poisson distribution with parameter one. By the central limit theorem, we find that the limit of $f_+(n)$ and of $(f_-(n))$ is $1/2$.